Álgebra lineal Ejemplos

$- 2 a + 3 b = 15$ , $\frac{4}{3} a + 2 = - 2 b$

Paso 1

Move variables to the left and constant terms to the right.

Toca para ver más pasos...

Paso 1.1

Suma $2 b$ a ambos lados de la ecuación.

$- 2 a + 3 b = 15$

$\frac{4}{3} a + 2 + 2 b = 0$

Paso 1.2

Combina $\frac{4}{3}$ y $a$ .

$- 2 a + 3 b = 15$

$\frac{4 a}{3} + 2 + 2 b = 0$

Paso 1.3

Resta $2$ de ambos lados de la ecuación.

$- 2 a + 3 b = 15$

$\frac{4 a}{3} + 2 b = - 2$

Paso 1.4

Reordena los términos.

$- 2 a + 3 b = 15$

$\frac{4}{3} a + 2 b = - 2$

$- 2 a + 3 b = 15$

$\frac{4}{3} a + 2 b = - 2$

Paso 2

Write the system as a matrix.

$[\begin{array}{c} - 2 & 3 & 15 \\ \frac{4}{3} & 2 & - 2 \end{array}]$

Paso 3

Obtén la forma escalonada reducida por filas.

Toca para ver más pasos...

Paso 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

Toca para ver más pasos...

Paso 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 3 & - \frac{1}{2} \cdot 15 \\ \frac{4}{3} & 2 & - 2 \end{array}]$

Paso 3.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ \frac{4}{3} & 2 & - 2 \end{array}]$

Paso 3.2

Perform the row operation $R_{2} = R_{2} - \frac{4}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

Toca para ver más pasos...

Paso 3.2.1

Perform the row operation $R_{2} = R_{2} - \frac{4}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ \frac{4}{3} - \frac{4}{3} \cdot 1 & 2 - \frac{4}{3} (- \frac{3}{2}) & - 2 - \frac{4}{3} (- \frac{15}{2}) \end{array}]$

Paso 3.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ 0 & 4 & 8 \end{array}]$

Paso 3.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

Toca para ver más pasos...

Paso 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ \frac{0}{4} & \frac{4}{4} & \frac{8}{4} \end{array}]$

Paso 3.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & - \frac{15}{2} \\ 0 & 1 & 2 \end{array}]$

Paso 3.4

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

Toca para ver más pasos...

Paso 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & - \frac{15}{2} + \frac{3}{2} \cdot 2 \\ 0 & 1 & 2 \end{array}]$

Paso 3.4.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{9}{2} \\ 0 & 1 & 2 \end{array}]$

Paso 4

Use the result matrix to declare the final solution to the system of equations.

$a = - \frac{9}{2}$

$b = 2$

Paso 5

The solution is the set of ordered pairs that make the system true.

$(- \frac{9}{2}, 2)$